$$
\begin{aligned}
& \text { (c) } f(x)=x^3-2 x^2-5 \text {, } \\
& \text { then } f(A)=A^3-2 A^2-5 I \\
& A=\left[\begin{array}{cc}
2 & -5 \\
3 & 1
\end{array}\right] \text {, then } \\
& A^2=\left[\begin{array}{cc}
2 & -5 \\
3 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & -5 \\
3 & 1
\end{array}\right]=\left[\begin{array}{cc}
-11 & -15 \\
9 & -14
\end{array}\right] \\
& \text { and } A^3=A \cdot A^2=\left[\begin{array}{cc}
2 & -5 \\
3 & 1
\end{array}\right]\left[\begin{array}{cc}
-11 & -15 \\
9 & -14
\end{array}\right] \\
& =\left[\begin{array}{cc}
-67 & 40 \\
-24 & -59
\end{array}\right] \\
&
\end{aligned}
$$

Use value of $A^2, A^3$ and $I$ in Eq. (i), we get
$$
\begin{aligned}
f(A) & =\left[\begin{array}{cc}
-67 & 40 \\
-24 & -59
\end{array}\right]-2\left[\begin{array}{cc}
-11 & -15 \\
9 & -14
\end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
-67 & 40 \\
-24 & -59
\end{array}\right]-\left[\begin{array}{cc}
-22 & -30 \\
18 & -28
\end{array}\right]-\left[\begin{array}{cc}
5 & 0 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
-50 & 70 \\
-42 & -36
\end{array}\right]
\end{aligned}
$$